help-17

[rapid swift]

you can add a 10-cent coin to produce an arrangement for n

[sullen rock]

Okay. I think I am getting lost when using inductive hypothesis, I need to prove that the statement is true for n + 1. Like 5(n + 1) cents

Is it just using that N > 4?

[rapid swift]

no

strong induction

is where you assume it's true from 4 through n and then prove it for n+1

[sullen rock]

Right okay.

[rapid swift]

so suppose you need to find an arrangement that makes 5(n+1) cents

I said that the intuition is that for any N >= 6, you can find an arrangement for it by finding an arrangement for N-2

and then adding a dime to it

so how would you find an arrangement for n+1?

[sullen rock]

Could you add a 25 cent coin?

[rapid swift]

no

just use what I said

you can, but there's absolutely no need to

[sullen rock]

Gonna be honest I legit so lost

I have 0 idea what to do

[rapid swift]

let's consider the case where n = 6

[sullen rock]

Ok.

[rapid swift]

you want me to prove that I can come up with an arrangement of coins to make 30 cents, right?

[sullen rock]

Yes.

[rapid swift]

my idea is to find an arrangement for n-2 = 4

and add a 10-cent coin to it

does that make sense?

by the strong induction hypothesis, I assume I can find an arrangement for anything less than 6, but at least 4

[sullen rock]

Ok

So

I think I keep looking a n-2=4 And not understanding where that number comes from or what it means

[rapid swift]

name any n

and I'll prove there's an arrangement by strong induction

[sullen rock]

n = 8

[rapid swift]

so strong induction says that we know we can find an arrangement of coins for 20 cents, 25 cents, 30 cents, and 35 cents, correct?

and your task is for me to prove it for n=8, i.e. 40 cents, right?

[sullen rock]

I suppose?

[rapid swift]

you suppose or you're sure?

[sullen rock]

Sure

[rapid swift]

the technique here

is I'm going to take the arrangement for 10 cents less

and just add a 10-cent coin

that corresponds to the arrangement for n-2

so I'm going to take the arrangement that's produced for n = 6, i.e. 30 cents, and I'm going to add a 10-cent coin to it

and now it's an arrangement of coins that adds up to 40 cents

[sullen rock]

Oh, Okay

[rapid swift]

do you see why this strategy works for any n >= 6?

[sullen rock]

Yes.

[rapid swift]

but do you see why it won't work for something like n = 5?

[sullen rock]

Yes

[rapid swift]

that's why we have to prove the base case for n = 4 and n = 5 separately

now how you formalize this is a little different

you want to prove it for n+1, assuming you can do it for 4, 5, ..., n

[sullen rock]

Right okay, So from the value 4 to the value of n

[rapid swift]

yes

you want to find an arrangement for n+1, given that you have arrangements for all of the rest below it

how might you do that, keeping in mind the strategy I gave?

[sullen rock]

Hmm

And this is for 5(n+1) Right?

[rapid swift]

this is for the statement that corresponds to n+1, which is that you can make change for 5(n+1) cents

[sullen rock]

Im not sure what to do from here

Im confused again already

[rapid swift]

confused about what

do you understand the task at hand?

[sullen rock]

Ok the strategy you gave me was adding 10 cents from doing n - 2

[rapid swift]

okay let's start over

very logically

[sullen rock]

Maybe I dont understand the task at hand

[rapid swift]

You are asked to prove the statement P(n) for all n >= 4

do you know what P(n) is?

[sullen rock]

Is it just 5n?

[rapid swift]

no

it is the statement that you can make change of 5n cents with 10-cent (hereafter referred to as dimes) and 25-cent (hereafter referred to as quarters) coins

[sullen rock]

ok

[rapid swift]

strong induction states that you can show that it's true for some base cases P(4), P(5), and then if you can show:
all of P(4), ..., P(n) is true IMPLIES P(n+1) is true
then you have shown that P(n) is true for all n >= 4

does that make sense?

we're making a logical argument here

not just writing shit down and hoping a professor accepts it and gives you a good grade

[sullen rock]

Okay, Yes

That makes sense

[rapid swift]

we've proved that P(4) and P(5) are true by exhibiting combinations of dimes and quarters that make 5(4) cents and 5(5) cents, correct?

[sullen rock]

Yes which is 5(4) = 2 dimes and 5(5) is a quarter

[rapid swift]

so what remains to show is the inductive case

i.e.

knowing that all of P(4), ..., P(n) is true means that we can also conclude that P(n+1) is true

[sullen rock]

Alright. Do you have to show what P(n+1) is or its just already implied it is true

[rapid swift]

you have to SHOW it's true, assuming you know that P(4), ..., P(n) is true

that's strong induction

if it's already implied that P(n+1) is true, then literally every single statement you can make out of your mouth is true

[sullen rock]

Okay

[rapid swift]

there's no implications here; I'm literally giving a proof that's based solely on logic and induction

so, using the definition of P, what we're really asked to show is

given that we can find combinations of dimes and quarters for 20, 25, 30, ..., 5(n-1), 5n cents, we can find a combination of dimes and quarters to get 5(n+1) cents

correct?

[sullen rock]

Correct

[rapid swift]

can you think of a way to do that, given that I told you the strategy before?

[sullen rock]

Please just refresh the "strategy" is. Are you talking about n - 2? Just want to make sure I am using the correct thing

[rapid swift]

yes the strategy is

for example, if you need to find a combination of coins that makes 90 cents

you can find one that makes 80 cents and then add a dime to it

[sullen rock]

Ok yes. So n = 16 5(16) = 80 + dime

To get the 90

I understand n - 2 now then adding the dime

[rapid swift]

now suppose you need to make 5(n+1) cents

how would you go about that?

[sullen rock]

n being any number I can think of but plus one?

5(8 - 1) 5(7) 35 cents? So I need a quarter and a dime

I feel like that is wrong considering I am trying to do the n - 2 strategy?

Oh wait

5(8 +1), 5(9)

which is 45

quarter and two dimes

Still feels like im off

[rapid swift]

yeah I have no idea what you're saying

[sullen rock]

Ok

[rapid swift]

I don't understand what's hanging you up at all

[sullen rock]

n + 1 is hanging me up

what + 1 does

[rapid swift]

n+1 is just a number

what's special about it

[sullen rock]

oh

[rapid swift]

if you have a number n

then you also have numbers n + 1

n+2

n+3

addition exists

the inductive case is

for any n, you need to show that P(4), ... P(n) implies P(n+1)

this is pretty simple logic, so I don't know what's going on

[sullen rock]

Yeah I am just not getting it

[rapid swift]

you're honestly telling me

that if you know how to make 20 cents, 25 cents, 30 cents, 35 cents, 40 cents

you can't figure out how to make 45 cents?

[sullen rock]

I can make 45 cents

one quarter and two dimes

[rapid swift]

you're honestly telling me that if you know how to make 20 cents, 25 cents, 30 cents, ..., 1585 cents, 1590 cents, you can't figure out how to make 1595 cents?

like ffs USE THE FACT THAT YOU CAN MAKE THE LOWER CHANGE AMOUNTS

[sullen rock]

yeah

I can figure that out

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[pure frost]

can someone explain how i would get the equation for a graph knowing the amplitude and period of the function (in terms of pi)

[vocal sleet]

@pure frost Has your question been resolved?

[sly sierra]

can you show a picture of an example?

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[vast shale]

How can I solve this application of a quadratic equation?
I don’t even know where to start.

[rugged orchid]

Start with (pipe1 + pipe2)*2 hours = 1 full tank

[vast shale]

What’s the next step?

@rugged orchid

[rugged orchid]

What’s the next line of the question

[vast shale]

Sorry?

[rugged orchid]

We just represented the first line of this question with math

Do the same for the second line

“One of the pipes used alone takes 3 hours longer than the other to fill the tank”

[vast shale]

P1=p2+3

[rugged orchid]

That says the speed of pipe 1 equals the speed of pipe 2 plus 3

[vast shale]

Isn’t that true ?

[rugged orchid]

What is 3

[vast shale]

One pipe takes 3 hours longer

[rugged orchid]

So the pipe1 and pipe2 represent speed

How can you add hours onto speed

Can you add 1 litre of water to 1 metre?

,rotate

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[woven sluice]

I have to use the shell method to find the area rotated about x=-1
My radius and height are (4-2x)(2-x) but im not sure if its correct

[woven sluice]

y=4-2x

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[woven sluice]

<@&286206848099549185>

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[woven sluice]

<@&286206848099549185>

is the radius 1+x?

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[manic edge]

.reopen

tell me definition of area of a figure

[lusty fox]

are you asking what is area

[paper depot]

in full generality?

[manic edge]

nope just simply tell me definition of " Area of a figure "

in sense

[lusty fox]

what

isnt that the same thing

[manic edge]

idk

@paper depot @lusty fox ?

[paper depot]

is this what you were looking for?

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[dusty nova]

Is this legal? Feels wrong to put the c constant affecting x inside the ln

[river minnow]

It's fine to do that

[dusty nova]

Cool, thanks. Just felt a bit weird putting it in there

.close

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[tranquil bison]

Hello, can someone explain more about sets and venn diagrams please?

[narrow light]

What do you know so far?

[tranquil bison]

i only know some symbols and their meanings

like union

and intercept

[narrow light]

But you know what a set is?

[tranquil bison]

yes

[narrow light]

So what do you want to know about?

[tranquil bison]

how to read the venn diagrams please should i send an example??

[narrow light]

Sure

[tranquil bison]

for the 2nd picture question a , how is it possible to get 5.7?

[narrow light]

Because the upside down "u" means the intersection

So it is the elements which are in both sets

[tranquil bison]

ohh thank you

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[rapid shore]

I have replicated an experiment, where 2 groups of participants estimated products of equations. In order to test the significance I am using a Mann-Whitney U Test but I dont really know why. Could someone help explain why this test is appropriate

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[vast shale]

S2

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[rapid shore]

<@&286206848099549185>

[crisp sedge]

?

[rapid shore]

.close

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[vernal harbor]

Ice cream is packaged in cylindrical gallon tubs. A tub of ice cream has a total surface area of 232.36 square inches.

If the diameter of the tub is 8 inches, what is its height? Use π = 3.14.

6.75 inches
5.25 inches
3.375 inches
2.625 inches

[eager walrus]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[vernal harbor]

1-2

actually i kinda have an answer but im not sure

i think its 5.25 inches

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[keen juniper]

i have to find the eigenvalues and eigenvectors of this matrix

[keen juniper]

i found i of order 2 and 1 for eigenvalues

and (1 0 0) for eigenvectors for i and ((-2c+1+i)/2 -1+i 1)

is this correct?

[wraith fable]

👍

you can always verify your results by multiplying the matrix with the eigenvector and see if you got it right

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[proven wyvern]

Hi I don’t understand how to do this question can anyone help me figure it out please

[proven wyvern]

I’ve tried to work out the angle at the circumference and double it like I was taught but it keep coming up as wrong

Here is my working out the working out isn’t the same as the question because it changed once I got it wrong

<@&286206848099549185> I know it hasn’t been 15 min but I need to get this done and go to sleep soon I’m sorry for any inconvenience

[rapid swift]

,rotate cw

[rapid swift]

which of the figures is for this problem?

[open otter]

[proven wyvern]

I need to figure out X

[proven wyvern]

On this ^

[rapid swift]

yeah I'm aware of that

[open otter]

I don`t know

[rapid swift]

you haven't told me which of the figures you've drawn corresponds to this problem

because I see four different figures, all with different angle measurements

[proven wyvern]

Ohh I understand now@

Umm the very top left one is the one I have been using

[rapid swift]

this one?

[proven wyvern]

Yes

I tried to figure out all the angles in those seperate shapes

As I know each shape adds up@to 360 degrees

[rapid swift]

your figure is drawn slightly incorrectly

[proven wyvern]

Oh how so ?

[rapid swift]

the orange thing is supposed to be bent the other way

[proven wyvern]

Ohhh I see

[rapid swift]

but regardless of that, it's usable

[proven wyvern]

I just don’t understand

My teacher isn’t very good when it comes to helping students

[rapid swift]

try to use the 80 degree angle to compute the measure of the red angle

[proven wyvern]

Ok thank you ill@try that

.close

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[crystal slate]

First photo is the question second is the answer booklet
How did they go from 1/2 - 1/(2^-x +1) = 1/2 - 2^x/(1+2^x) = -1/2 +1/(1+2^x)
And if it’s wrong what would be the correct way (this answer book has been wrong in the past)

[crystal slate]

<@&286206848099549185>

[upper comet]

To prove that a function is odd, we need to show that it satisfies the property $f(-x) = -f(x)$ for all $x$ in the domain of the function.

Given the function $f(x):$

$$
f(x) = \frac{1}{2} - \frac{1}{2^{-x}+1}
\
= \frac{1}{2} - (2^{-x}+1^{-1})
$$

Let's start by evaluating $f(-x)$:

$$
f(-x) = \frac{1}{2} - \left(2^{-(-x)} + 1^{-1}\right)
$$

We know that $2^{-(-x)}$ is equal to $2^x$, and $1^{-1}$ is equal to $1$, so we can simplify this expression:

$$
f(-x) = \frac{1}{2} - \left(2^x + 1\right)
$$

We recall that we can write $f(x)$ as:

$$
f(x) = \frac{1}{2} - \left(2^{-x} + 1^{-1}\right)
$$

Therefore, we know that $2^{-x}$ is equal to $1/2^x$, and $1^{-1}$ is equal to $1$, so we can simplify thus:

$$
f(x) = \frac{1}{2} - \left(\frac{1}{2^x} + 1\right)
$$

... et cetera

[twin meteor]

gnu

[lusty fox]

did you write all that

!nosols

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[upper comet]

It only adds detail to the answers they have access to in their answer key

it's a partial solution :)

[crystal slate]

Alright thank you! My math class just ended to I’m going to screenshot it and look through later lol

[upper comet]

No problem!

[crystal slate]

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[tame zephyr]

(Sorry for the vertical picture)

[karmic imp]

,rotate

[tame zephyr]

Can anyone help me solve this? I can only get a negative value for x

<@&286206848099549185>

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@tame zephyr Has your question been resolved?

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@tame zephyr Has your question been resolved?

[fluid obsidian]

wth

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[royal grove]

this is wrong

this all just cancels out and you'll get 1 = 1

its the same thing in both numerator and denominator

take the dot product of a and b, that is
(3i + 4j).(i + xj) = 3 + 4x

the denominator is correct

so the final expression should come out to be

[tame zephyr]

Why does that equal to 3+4x?

[royal grove]

take the dot product

(3i . i) + (3i . xj) + (4j . i) + (4j . xj)

i . i = j . j = 1
i . j = j . i =0

@tame zephyr

[tame zephyr]

I'm trying it on paper

[royal grove]

👍

[tame zephyr]

So if I were to make that into x= it would be like this right?

Forgot to add the -9 there

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[cunning veldt]

please help im stuck trying to solve for h

[vocal sleet]

Please don't occupy multiple help channels.

[cunning veldt]

.close

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[rapid berry]

can someone help me

[paper depot]

http://dontasktoask.com/

[rapid berry]

huh

Uh

[river minnow]

Post the question, don't ask for the permission to do that next time

[rapid berry]

oh

[river minnow]

That's what "Don't ask to ask, just ask" means

[rapid berry]

ohhhh

Okay next time I'll send it

,close

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[river minnow]

Don't you have a question rn?

[rapid berry]

uh

kinda

[river minnow]

.reopen

[vocal sleet]

✅

[rapid berry]

Wait uhhh

I'll send my question

hold on

Uh

actually uhhh

Nvm

Uhm

[vocal sleet]

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[late sky]

Does Bolzano weistrass apply to finite sequences?

[late sky]

I don’t think so, take 1,2,3,4,5 for example

[cinder temple]

hi

[hybrid flicker]

a finite sequence converges by definition

even though the notion of convergence for finite sequences is a bit shaky

in order to apply bolzano weierstrass, you need an infinite sequence

[cinder temple]

It should work from my research

[hybrid flicker]

?

[cinder temple]

Well'

[late sky]

How,could you elaborate? Using the standard epsilon definition of convergence how does that apply to finite sequences like 1,2,3,4,5

[hybrid flicker]

well what would be the limit of a finite sequence?

[late sky]

It’ll be 5

But that doesn’t really have the same notion of convergence as infinite sequences do?

[paper depot]

you're right, there is no notion of convergence for a finite sequence.

[late sky]

For this proof

They’re choosing xn such that f(xn)>n

But how can you know for sure that this xn exists in the interval of [a,b]

[hybrid flicker]

first of all, f is only defined in [a,b]

so it goes without saying that xn such that f(xn) > n would be in [a,b]

[late sky]

But how do you know this xn exists?

[hybrid flicker]

because f is not bounded

[late sky]

Oh

Thank you for the clarification

[hybrid flicker]

(at least we suppose f is not bounded)

[late sky]

Yes, it’s a contradiction proof but I overlooked the initial few lines

Thank you

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[hybrid flicker]

no problem

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[vocal sleet]

@tacit aspen Has your question been resolved?

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[noble citrus]

use square completion to find the center and halv-axis to (x,y,z):2x^2-4x+5y^2+10y+z^2-4z-14=0

[noble citrus]

here's my attempt. Does it look correct?

[vocal sleet]

@noble citrus Has your question been resolved?

[hybrid flicker]

You underline "-1", "-2" and "-1" but you reverse their signs?

also, it would be better to get them out of the parenthesis multiplied by the coefficient outside of the parenthesis

[noble citrus]

Yeah, in order to square complete for x, y, z I had to add a constant, and at the same time subtract the constant, in order to keep it "balanced" I underlined the subtracted constants and moved them to the other side of the equation causing them to change signs.

Oh, I see an error here. Before moving them to the other side of the equation I first have to multiply them by the coefficient before the parenthesis

[hybrid flicker]

The problem is, you already "balanced" the equation by subtracting these constants

They won't be brought to the other side, everything is on the Left hand side

[noble citrus]

well, in that case I must have misunderstood my lecturer, here's a snippet from my lecture notes:

[hybrid flicker]

If you take a look at this snippet, looking at 9(x²-2x+1) +...

Since the "1" has been added, you need to subtract it in return

[hybrid flicker]

Now let's take a look at your snippet in x here

2(x²-2x+1-1) ...

Since the "+1" was added, in return you subtracted "-1" to bring back balance

There was already balance, you didn't need to change the sign again to throw it off balance again

[noble citrus]

Oh okay, so it should look like this for example: 2(x^2-2x+1) and then I multiply 2 * 1 and move it to the other right side?

if not, then I have no idea what's going on in the right side of my lecturers example on the snippet.

[hybrid flicker]

Yes that's it

Or, you can switch the sign and keep it on the same side

[noble citrus]

Alright, I got it, thanks!! 🙂

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[vast shale]

Hello guys!

[vast shale]

i have this question regarding max cannibalisation

here are the tables of data for reference:

Okay so i am *pretty confused *on one thing here

from what i know, this simply should be

(Smarthub Refrigerator's Contribution Margin * Smarthub Refrigerator's Sales - Fixed costs) / Standard No-Frost Refrigerator's Contribution Margin

but seemingly that is not correct?

[real gale]

"max cannibalisation"

[vast shale]

yes thats a concept in accounting

it is basically what the maximum amount of sales can you lose from an old product to a new product without resulting in a lesser total profit than if you had only sold the old product

what the question is asking, basically

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

@vast shale Has your question been resolved?

[open cave]

I KNOW WHAT THAT IS

i remember

i told you

ehhe

[open cave]

my brother in christ, BECR (max cannibalization rate) = Total contribution per new product / total contribution per existing product

[vast shale]

that doesn't answer my question at all

[open cave]

hmm

not obvious already? ok

[vast shale]

sigh dude whatever

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[open cave]

astagfrullah

patience is a virtue, grasshopper

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[radiant mountain]

i need help so how we solve -15/8multiply-12/5multiply2/-3

[rugged vortex]

Please send a screenshot or say it in a clearly way

[open cave]

ya send it in a clear way

[open cave]

however i think i know what you mean

????????

[radiant mountain]

yea

[open cave]

hmm

cancel out common factors

[radiant mountain]

i have to calculate by detailling

[open cave]

bro

just cancel

[radiant mountain]

cancel what

[rugged vortex]

Prime factorize each number down

And then cancel out like factors

[radiant mountain]

ok

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[half summit]

is my answer right

[vast shale]

Indeed that is by how much the area will change

[half summit]

awesome, thank you

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[scenic dagger]

Help

[scenic dagger]

How do I do this

[hushed pewter]

In the graph, the purple lines represent Olivia's distance from her home at a given time

[scenic dagger]

But how do I get d

[hushed pewter]

Here at the beginning is just before she starts her walk. No time has passed yet (that is, 0 minutes have passed), and she is at her home (she has 0 distance from her home)

And here is where 10 minutes have passed

[scenic dagger]

Ok

[hushed pewter]

So d is how far she walked in 10 minutes

[scenic dagger]

How do I get it

[hushed pewter]

She stays at her friend's house for 30 minutes, so the height of the purple line stays constant between the timespan of 10 minutes to 40 minutes (30 minutes elapsing)

It's given in the question

[scenic dagger]

Ahh I see

Thanks

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[scenic dagger]

. Close

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[edgy dagger]

Hi
How to put into vertex form and tell if its being stretch /compress ?

[vast shale]

,tex .cts

[twin meteor]

Toby

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[lofty canyon]

468x + 222y = 6
The graph of this should give me ordered pairs of x and y that satisfy the equation, but when i plot the graph, at x = -9, y is not an integer but while solving, y = 19. Can someone explain why this is happening?

[sharp lynx]

when i plot it, it works fine

perhaps show what you're plotting with

[lofty canyon]

man, i was looking at the wrong axis, guess the sleep deprivation has started to kick in

im sorry for bothering you, thanks

man i feel like such a dum dum

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[lean osprey]

Could someone explain the second step in the first derivative part?

[sharp lynx]

looks like they factored out a common term of (x-a)^[m-1]*(x-b)^[n-1]

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[radiant sapphire]

Question: When is the particle moving to left?

[radiant sapphire]

Since v(t) can't be 0, I don't think I can use inequality.

[sharp lynx]

why can't v(t) be 0?

[vast shale]

why cant v(t) be 0?

[sharp lynx]

sniped

[vast shale]

[radiant sapphire]

Because (1-t^2) can't be zero

[sharp lynx]

sure it can

[radiant sapphire]

Oh wait

It can

[sharp lynx]

the other one can't be 0

[radiant sapphire]

How would you draw the inequality table?

[sharp lynx]

well your other term is always positive, so you'd draw it in the exact same way you'd draw it for (1-t^2)

[radiant sapphire]

Like this?

[sharp lynx]

i'm not sure how to parse this

but remember (1-t^2) = 0 for two different values of t

[radiant sapphire]

Oh not only 1?

[sharp lynx]

not only 1

[radiant sapphire]

Uh..

What is the other one

[sharp lynx]

,w plot 1-x^2

[twin meteor]

Results provided by WolframAlpha

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Requested by Steakanator#3427

[radiant sapphire]

Ohh

I get it

So -1 and 1

[sharp lynx]

actually wait i forgot you're measuring time

ignore me, in this case it would only be 1

[radiant sapphire]

Ah okay

[radiant sapphire]

I got the answer, thank you for the help!

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[dark cape]

smh.........

[tall ore]

Do you know what rules are used when you have 2 sides and angle

[vocal sleet]

@dark cape Has your question been resolved?

[dark cape]

no

[cinder temple]

ok

try the cos rule

@dark cape

Helloooo?

[dark cape]

oop

okay

thanks

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[glad pendant]

I was just wondering why in the answers, in the synthetic division he subtracted rather than adding

I did this for the synthetic division

but they did this

and im not sure why

[dusk bay]

your way works

but you have to add

[glad pendant]

my way i didnt get the same answer

[dusk bay]

because you subtracted

what's 9+2 vs 9-2

the way we do synthetic division is we take whatever x value makes the expression = 0 (which is what you did)

[glad pendant]

ohhhhh

[dusk bay]

then you add the numbers

[glad pendant]

right right right

sorry

thank you so much

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[split delta]

How do we go from the first step to the second step?

[lime gorge]

a^2 - b^2

= (a+b)(a-b)

U see it? @split delta

[split delta]

no

can you explicitly do it?

with the terms f and g?

[lime gorge]

So let (5-f) = a and let (5-g) = b

Try going from there

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[river kettle]

How would I deal with this?

I know a is (2,8) and that it’s a right angle triangle

I don’t know what to do after thoufh

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[vast shale]

What do I do next

Am completing the swuare

[floral pike]

iirc, setting equal to zero and dividing out a messes you up

[vast shale]

Am asking because I have a perfect square in the denominator

[vast shale]

?

[floral pike]

Factor it out instead

[vast shale]

iirc?

[halcyon compass]

what's the original question here

iirc stands for if i recall correctly

[floral pike]

Oh nvmd

[halcyon compass]

or remember

[vast shale]

Original question

[halcyon compass]

and you need the roots?

[vast shale]

Correct

But I need solve by completing the square

[halcyon compass]

just use second power equation

ah okay

[floral pike]

Just subtractive 1/2 from both

Those are your x values

The solutions

Just solve for x

[dull bear]

[should be 1/4 in that line written]

[vast shale]

I am confused

I’m asking

After I apply the square root property

What do I do next

Since there is a perfect square in the denom

[floral pike]

Oh, well ok

Yes, I see the typo

[vast shale]

Typo?

See, I messed up because I learnt this in decimals, but now I realized that I should be doing it with fractions

[floral pike]

The constant inside the root is 1/4, but you wrote 1/2 on the next line

I am acknowledging you catching your own mistake there

If your goal is to find the roots, you just need to get x by itself now

...by subtracting 1/4 from both sides of the equation

Is that all you needed?

[vast shale]

No

I am very confused now

What I’m saying is

I messed up from the beginning when I was learning this unit, because on the 3rd step where you are supposed to divide the coefficient of x by 2 and square it. I learned this by doing that by just dividing it my 2 in my calculator but now the answer key is showing a fraction and idk how to write it as a fraction so I’m fucked

So, I need to learn how to do that step in fractions

[vast shale]

I’m not sure what you mean by this

[floral pike]

Oh! Lol

That's just a basic law of exponents

You did fine squaring the fraction on this paper...?

$\left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2}$

[twin meteor]

Disorganized

[floral pike]

@vast shale feel better?

@vast shale you realize that you have a mistake on the last line and that looks like the only problem?

[vast shale]

Yes I understand that

[vast shale]

Could u underline my mistake?

[floral pike]

You said yourself that thr 1/2 should be 1/4

Because that's what's on the line above it

Not 1/2

That was a typo

You were supposed to copy what was inside the parentheses

Look at your last two lines

It should be x + 1/4, not x + 1/2

@vast shale

Are we done?

[vast shale]

OHHHH

OKAY I SEE

But no that wasn’t my problem LOL

[floral pike]

Well besides that your paper doesn't seem to have a problem

[vast shale]

Don’t I have to get rid of that perfect square that is present in the denominator ?

The 16s?

[floral pike]

No

Oh

Yes

Sqrt of 16 is 4

Done.

Sqrt(17)/4

[vast shale]

So that would be my answer?

[floral pike]

Also, if you wanna be hard on yourself, the second-to-the-last (penultimate) line should have a +- symbol in front of the sqrt on the right. One side must have this, and since your subject is on the left (the x), do the right

[vast shale]

The answer key says -1+_root17/4

[floral pike]

You still didn't fix the 1/2

Whatever lemme check this on a napkin rq

[vast shale]

Oh mb, but that doesn’t make a diff my answr would still be wrong

[frozen bobcat]

your 1/4 became a 1/2 when you square rooted, it shouldn't have

[floral pike]

Ooh, it's gonna take a min to upload this image

But you also have am error with thr +- symbol

[floral pike]

Finally got wifi

[vast shale]

It’s still different from the final answer

In the key

[floral pike]

^you said you understood that part

Hmmm....

[vast shale]

The key says -1+root17/4 and -1-root17/4

[floral pike]

Ohhhhh

It's the Factor of a

Didn't account for it

[frozen bobcat]

$\frac{-1\pm\sqrt{17}}{4}$ or $-1 \pm \frac{\sqrt{17}}{4}$

[twin meteor]

Zybikron

[floral pike]

,rotate

[floral pike]

Nvrmnd it's the same, I'm losing it

[vocal sleet]

@vast shale Has your question been resolved?

[floral pike]

ok

@vast shale

here's an important detail I forgot to bring up

notice that if you distribute

$\left(x - \frac{-1 - \sqrt{17}}{4}\right)\left(x - \frac{-1 + \sqrt{17}}{4}\right)$

[twin meteor]

Disorganized

[floral pike]

you will get

$x^2 + \frac{1}{2}x - 1$

[twin meteor]

Disorganized

[floral pike]

multiply this through by a=2, and you get what you started with.

stationary points on a graph (the zeros) don't move when you multiply through by a scalar (a constant value)

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[oak owl]

what does this mean anyone ?

[desert hornet]

do you know what the domain of a function is?

[oak owl]

not really

since its not my main language so Im not so sủe

sure*

also can someone help with this

[vocal sleet]

@oak owl Has your question been resolved?

[oak owl]

<@&286206848099549185>

[lusty fox]

whats the question

[oak owl]

The first one is to find the domain of function

the second one is just solving it

[lusty fox]

oh i got it

factor them out

looks tricky but actually isnt

[oak owl]

the thing is how tho

also how do i find the domain of function

[lusty fox]

bro

!nosols

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[lusty fox]

delete that

[lusty fox]

thats for later

[lusty fox]

do you not know how to factor quadratics

[oak owl]

I kinda does

[river nebula]

oh oops - first time here.

got it

[lusty fox]

no worries

[lusty fox]

factor: x^2 + 4x + 4

[oak owl]

thats (x-2)^2

righjt ?

[lusty fox]

no

[oak owl]

wait

what does factoring means then

[lusty fox]

i mean youre close

expand out (x-2)^2

[oak owl]

ohhhh

wait

gimme a sec

[oak owl]

this doesnt expand much futher tho

[lusty fox]

it does

[lusty fox]

expand this out

to check your work

[oak owl]

Im still a bit confused since english is not my main language

So we don't have terms as factoring

[lusty fox]

we do

just expand out (x-2)^2

make it into a polynomial in normal form

[oak owl]

(x-2)(x-2)

[lusty fox]

expand that out

multiply the terms

[oak owl]

oh

its (x+2)^2

not x-2

thats right

[lusty fox]

yes

[oak owl]

my bad on that

read it a bit wrong there

oh so i just factor it out and delete the square root

then just do it

[lusty fox]

well no

you can only "delete" (we use the word "cancel" in math) the square root if the quadratic is a perfect square

[oak owl]

right

[lusty fox]

which just so happens to be the case here since the problem is meant to be relatively straightforward to solve

[oak owl]

yep got it

in this case we gonna end up with x-3 + x+1 = 6

and x=4

[river nebula]

👍🏻

[oak owl]

how about this ?

This is just a simplify question

[river nebula]

Yep so work on what’s in the brackets first then.

[oak owl]

I tried just mutiplying them in the brackets but didn't went so well

so it got really confusing after the mutiplication

[river nebula]

What do you have so far?

[oak owl]

ok so I went and mutiply everything in the brackets like 3 to the a^2+a/5 and 3a-2/a+1 to a^2+a/5

and it get confusing really quickly

[river nebula]

Ok. So you have 2 expressions now. Look at it one by one. Can you further simplify anything ? 😉

[oak owl]

I have 2 equation here

3a^2+3a/15

which can go down to a^2+a/5

if i divide it by 3

but prob not

[river nebula]

The denominator won’t change

[oak owl]

the other equation being 3a^3 +a^2-2a/5a+5

[river nebula]

Haha hold on on a sec. Let go back to the first one.

[oak owl]

okay

[river nebula]

If we take 3 times 1/5, we get 3/5 isn’t it?

[oak owl]

yes

oh yea

[river nebula]

So if you multiply 3 to that fraction the denominator won’t change haha

[oak owl]

thats where i went wrong

but still

[oak owl]

the second equation is this

[river nebula]

Ok now the second one. Before we expand the brackets. Are there stuffs that we can actually cancel off the numerator and the denominator?

Trying leaving them In their brackets and observe.

[oak owl]

Im seeing a^2+a and a+1

[river nebula]

Yeah can we do something to the a^2+a first?

[oak owl]

where a^2+a is just (a+1). a

[river nebula]

Yup 👍🏻 so what can we do?

[oak owl]

so we divide the second one to a ?

a/a to be exact

i think ?

[river nebula]

You see now the top and bottom Both have (a+1). What can we do?

[oak owl]

wait i got it wrong a bit

just take a out

i mean

then we cancel it out

[river nebula]

Yup

[oak owl]

so we get 3a-2 . a/5

[river nebula]

Yup

[oak owl]

and we get 5a/5 with the first equation

so its loga(a)

so its 1

okay

nice

[river nebula]

😁

[oak owl]

got ittt

okayy

wait how do i find the domain fuction ?

Im not great with math so if you can help me out it would be so much an appreciate

[river nebula]

Any example ?

[oak owl]

yup

here

I don't know how to

[whole oasis]

thing under the square root >= 0

because in real world you cannot take square root of negative numbers

[oak owl]

that is true

yes

[whole oasis]

hence for a)

[oak owl]

the remaining 2

tho

its

a bit more

[whole oasis]

you should do sinx >= 0

[oak owl]

so sin x =>0

f(x) = [0, infinite)

[whole oasis]

?

you're supposed to solve for x

aa, you're talking about the range

[oak owl]

nope its just the find the domain of the function

yyup

all these question is just domain of functions

[whole oasis]

but it's wrong

anyway

[oak owl]

hmm ?

[whole oasis]

,w plot sinx from -2pi to 2pi

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Modus#7591

[whole oasis]

look at this and try to determine where wave is above the x axis

(it's enough to analyze only one period)

[oak owl]

wait

what does domain of function actually means

its when the equations work right ?

[whole oasis]

it means all numbers 'x' you can plug into the function and get an output

[oak owl]

like the range where x is in for the equation to make sense

ok

[whole oasis]

it's domain

range is all possible values that function can achieve

[oak owl]

ok

[whole oasis]

(so when determining range we look at the y - axis)

[oak owl]

and when determining domain we look at x - axis ?

[whole oasis]

yes

but here we have only formula

[oak owl]

yes

[whole oasis]

and we can deduce that output exists when sin(x) >= 0

because of what I've said

[oak owl]

oh so we write in sin(x) >=0

[whole oasis]

yes, it's our condition

[oak owl]

not the range

[whole oasis]

but you should solve for x to get the domain

[oak owl]

yup

[whole oasis]

yeah, domain and not range

[oak owl]

sin(x) = 0

so i solve x like that ?

because with that x =>0

=0 i mean

is that correct ?

[vocal sleet]

@oak owl Has your question been resolved?

[vocal sleet]

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[hard raven]

help pleaase

[vocal sleet]

@hard raven Has your question been resolved?

[vocal sleet]

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[soft arch]

my question is basically

why do i have to do n ln (1+2/n)

instead of root test

why cant i root test this lol

<@&286206848099549185>

.close

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[carmine kettle]

Given this matrix, are there any matrices other than
B=[0 0]
[0 0]
such that AB=BA

[paper depot]

yes

[carmine kettle]

I tried to say that
B= [ a b ]
[ c d ]

[low lance]

Jus take B = [a b c d] and put the condition AB = BA u will get

[carmine kettle]

and solved for a,b,c,d

[low lance]

Yes

[carmine kettle]

only thing im getting is 0s

[paper depot]

show work

[carmine kettle]

I got 4 equations

give me a second

am I supposed to equate to 0 like I did

that's the part im unsure about

[paper depot]

yes, the zero matrix is the matrix with all entries equal to 0

[carmine kettle]

I see

[paper depot]

what do you get when you simplify each equation?

do not try to combine them, only simplify them in isolation

[carmine kettle]

everything just cancels out

alright give me a second

[paper depot]

it is not quite as sweeping as that.

[carmine kettle]

I just get c=0, d=a

[paper depot]

exactly.

so you get that the matrices which commute with $A$ are precisely those of the form $\bmqty{a & b \ 0 & a}$

[twin meteor]

Ann

[carmine kettle]

ohh

so I get like a general matrix

[carmine kettle]

so for every a,b here
AB=BA is true?

[paper depot]

yes, and for no others.

[carmine kettle]

I see

Thanks a bunch

.close

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[lament flicker]

why did he take x/2 as a usub?

and how can something like this come up in my mind in the test?

[crimson jetty]

isnt that just some bot

Factorize 4 out of the entire square root first

and u will come to the same conclusion

So factor 2 = sqrt(4) out

[lament flicker]

How do i do that?

oh nvm i got it

.close

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[deft sedge]

Q's

1. where is this sketchy formula coming from (first formula, top left)
2. Why is T = time ? shouldnt 5 be plugged into variable"t" instead?
3. why is exponent a fraction: t/T

[carmine leaf]

1. its exponential function using different base than e

2. its t/T in the exponent t is still time T is another constant

[deft sedge]

T= 5

[carmine leaf]

yes, so if t=5 that means you have 500 * 2 which is what you want right

for the initial value 500 to double in 5 years

[deft sedge]

shouldnt it be 5/T

not vice versa